Let $f$ be an analytic function on the open unit disc $D=\{z \in \mathbb{C} \mid |z|<1\}$, suppose it is continuous on the the closure $\bar{D}$, then is $f$ also analytic on $\bar{D}$?
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1$\begingroup$ Here is a counter-example. $\endgroup$– 23rdJun 5, 2013 at 1:27
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$\begingroup$ Great，thank you! I would like to consider it as an answer if you wrote in as an answer below. $\endgroup$– Li YutongJun 5, 2013 at 1:56
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1$\begingroup$ You are welcome! Actually there are more pathological examples: $f$ could be not analytic at any point in $\partial D$. I hope you or someone else would provide an answer with more elaborate discussion. $\endgroup$– 23rdJun 5, 2013 at 2:04
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$\begingroup$ math.stackexchange.com/q/390407/462 provides a counterexample. If $f$ where analytic on $\bar D$, its series would converge uniformly in a neighborhood of each point of $\partial D$ (and more). The given answer shows that we can have non-uniform convergence at every point of $\partial D$. $\endgroup$– Andrés E. CaicedoJun 5, 2013 at 2:54
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A simple counterexample is $f(z)=\sqrt{1-z}$, where the square root is understood as the principal value. This function has a continuous extension to the closed disk, but it is not Lipschitz continuous on it. If it were holomorphic on a domain containing $\overline{D}$, it would be Lipschitz continuous on $\overline{D}$, due to the derivative being bounded.